Question

Identical charges of magnitude Q are placed at (n-1) corners of a regular polygon of n sides each corner of the polygon is at a distance r from the centre. The field at the centre is what ?

Answer 1

answer

r(n−1)

Explanation:

Correct option is

Correct option isB

Correct option isBr(n−1)

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r(n−1)q

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r(n−1)q

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r(n−1)q

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r(n−1)q The electric filed is a vector quantity. So the electric field cancel each other for the charges of opposite corner of polygon. Only charge nq−(n−1)q=q will contribute the electric field at the center of polygon. thus, E=k

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r(n−1)q The electric filed is a vector quantity. So the electric field cancel each other for the charges of opposite corner of polygon. Only charge nq−(n−1)q=q will contribute the electric field at the center of polygon. thus, E=k r

Correct option isBr(n−1)The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to (n−1)q number of charges. i.e, V=k r(n−1)q The electric filed is a vector quantity. So the electric field cancel each other for the charges of opposite corner of polygon. Only charge nq−(n−1)q=q will contribute the electric field at the center of polygon. thus, E=k r 2